What is the slope of the line tangent to $f(x) = -x^{2}+x+3$ at $x = 3$ ?
Solution: The slope of the tangent line is $ \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$ $ = \lim_{h \to 0} \frac{(-(x+h)^{2}+x+h+3) - (-x^{2}+x+3)}{h}$ $ = \lim_{h \to 0} \frac{(-(x^{2}+2x h+h^{2})+x+h+3) - (-x^{2}+x+3)}{h}$ $ = \lim_{h \to 0} \frac{-x^{2}-2(x h)-h^{2}+x+h+3+x^{2}-x-3}{h}$ $ = \lim_{h \to 0} \frac{-2(x h)-h^{2}+h}{h}$ $ = \lim_{h \to 0} -2x-h+1$ $ = -2x+1$ $ = (-2)(3)+1$ $ = -5$